Just to give you a context (but not directly related to my question, so you can skip this paragraph and go to next one): I've been reading http://www.math.iit.edu/~fass/603_ch2.pdf for a characterization of completely monotone functions, and I see that Theorem 2.5.2 (Hausdorff-Bernstein-Widder theorem) states that a characterization of completely monotone function $\phi:[0,\infty)\to \mathbb{R}$ is equivalent to the Laplace transform of a finite non-negative Borel measure on $[0,\infty)$, i.e. equivalent to writing: $\phi(r)= \int_{0}^{\infty}e^{-rt}d\mu(t)$, where $\mu$ is a non-negative Borel measure on $[0,\infty)$. I'd like to find the non-negative Borel measure for the functions corresponding to Gaussian, as made explicit below.
My question is: Let $\alpha \geq 0$. Then what's a finite, non-negative Borel measure $\mu$ on $[0,\infty)$, satisfying: $\int_{0}^{\infty}e^{-rt}d\mu(t)=e^{-\alpha r} \forall r \in [0,\infty)$?
If we take: $d\mu(t):=\alpha dt$ on $[\alpha, \infty)$ and $0$ on $[0,\alpha)$, that satisfies the above equation, but $\mu$ isn't finite in this case.